Physics

Thermal Properties and Thermodynamics

380 Questions

Thermal properties and thermodynamics questions evaluate concepts of heat transfer, thermal efficiency, and temperature variations. Problems involve calculating heat content, conductivity, and the performance of heat engines. This subject is regularly tested in physics sections across multiple competitive platforms.

Heat transfer calculationsThermal efficiencyBlack body radiationTemperature variationsRefrigeration performance

Thermal Properties and Thermodynamics Questions

Multiple choice stefan's law black body radiation heat transfer thermal properties physics

A black body is heated from $27^oC  $ to $927^oC  $. The ratio of radiation emitted will be:

  1. $1 : 4$
  2. $1 : 8$
  3. $1 : 16$
  4. $1 : 256$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Energy radiated depends on the temperature of the body.
Stefan's law states that the total amount of energy radiated per second per unit area of a perfect black body is directly proportional to the fourth power of the absolute temperature of the surface of the body,ie,
$E\propto { T }^{ 4 }$
or $E=\sigma { T }^{ 4 }$
where $\sigma $ is Stefan's constant. It's value is $5.67\times { 10 }^{ -8 }W{ m }^{ -2 }{ K }^{ -4 }$
Here, ${ T } _{ 1 }=27+273=300K$
          ${ T } _{ 2 }=927+273=1200K$
$\therefore       \dfrac { { E } _{ 1 } }{ { E } _{ 2 } } ={ \left( \dfrac { 300 }{ 1200 }  \right)  }^{ 4 } = 1 : 256$

Multiple choice stefan's law black body radiation heat transfer thermal properties physics

Two bodies A and B of equal surface area have thermal emissivities of $0.01$ and $0.81$ respectively. The two bodies are radiating energy at the same rate. Maximum energy is radiated from the two bodies A and B at wavelengths $\lambda _A$, and $\lambda _B$ respectively. Difference in these two wavelengths is 1 $\mu$. If the temperature of the body A is $5802\  K$, then value of $\lambda _B$ is :

  1. $\dfrac{3}{2}\mu m$
  2. $1\mu m$
  3. $2 \mu m$
  4. $\dfrac{3}{4} \mu m$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We know that as per stephan's boltzman radiation law, $P\ \alpha\  \sigma A{ T }^{ 4 }$.
Since surface area of two bodies is same,

Therefore, ${ \sigma  } _{ A }{ T } _{ A }^{ 4 }={ \sigma  } _{ B }{ T } _{ B }^{ 4 }$. Hence, ${ T } _{ A }=3{ T } _{ B }$.

Now, as per Wein's displacement law, $\lambda T=constant=k$, $\lambda =\dfrac { k }{ T } $.

${ \lambda  } _{ B }-{ \lambda  } _{ A }=k(\dfrac { 1 }{ { T } _{ B } } -\dfrac { 1 }{ { T } _{ A } } )=k(\dfrac { 1 }{ { T } _{ B } } -\dfrac { 1 }{ 3{ T } _{ B } } )=\dfrac { 2k }{ 3{ T } _{ B } } =1\mu $


${ T } _{ B }=\dfrac { { T } _{ A } }{ 3 } =1934\ K$

Putting in the above equation to calculate $k$ and the solving for ${ \lambda  } _{ B }=\dfrac { k }{ { T } _{ B } } $, we get ${ \lambda  } _{ B }=1.5\mu m$.


Multiple choice stefan's law black body radiation heat transfer thermal properties physics

A black body at a high temperature $T$ radiates energy at the rate of $U\left( in\quad W/{ m }^{ 2 } \right) $. When the temperature falls to half (i.e $T/2$), the radiated energy $\left( in\quad W/{ m }^{ 2 } \right) $ will be

  1. $U/8$
  2. $U/16$
  3. $U/4$
  4. $U/2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

According to Stefan's law, rate of energy radiated by a black per unit area (inW/m2)(inW/m2) at temperature TT is given by
U=σT4...(i)U=σT4...(i)
when the temperature falls to half (i.e., T/2T/2) 
radiated energy (inW/m2)...(ii)(inW/m2)...(ii)
From (i) and (ii) we get
U′U=(12)4=116U′U=(12)4=116
or U′=U16

Multiple choice stefan's law black body radiation heat transfer thermal properties physics

 If the radius of a star is R and it acts as a black body, what would be the temperature of the star, in which the rate of energy production is 0? (a stands for Stefan's constant.)

  1. $

    \left(\frac{4 \pi R^{2} Q}{\sigma}\right)^{1 / 4}

    $
  2. $

    \left(\frac{Q}{4 \pi R^{2} \sigma}\right)^{1 / 4}

    $
  3. $

    \frac{Q}{4 \pi R^{2} \sigma}

    $
  4. $

    \left(\frac{Q}{4 \pi R^{2} \sigma}\right)^{-1 / 2}

    $
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice stefan's law black body radiation heat transfer thermal properties physics

Assuming the Sun to be a spherical body of radius $R$ at a temperature of $T\ K$. Evaluate the intensity of radiant power, incident on Earth, at a distance $r$ from the Sun where $r _{0}$ is the radius of the Earth and $\sigma$ is Stefan's constant :

  1. $\dfrac{R^{2}\sigma T^{4} }{r^{2}}$
  2. $\dfrac{4\pi ^{2}R^{2}\sigma T^{4}}{r^{2}}$
  3. $\dfrac{\pi ^{2}R^{2}\sigma T^{4}}{r^{2}}$
  4. $\dfrac{\pi ^{2}R^{2}\sigma T^{4}}{4\pi r^{2}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Total power radiated by the sun

 $=\sigma { T }^{ 4 }\times 4\pi { R }^{ 2 }$

The intensity of power at earth surface

$=\cfrac{\sigma { T }^{ 4 }\times 4\pi { R }^{ 2 }}{4\pi { r }^{ 2 }} \\=\cfrac{\sigma { T }^{ 4 } { R }^{ 2 }}{{ r }^{ 2 }}$
Multiple choice stefan's law black body radiation heat transfer thermal properties physics

The rectangular surface of area $8 cm \times 4 cm$ of a black body at a temperature of $127^0C$ emits energy at the rate of $E$ per second. If both length and breadth of the surface are reduced to half of its initial value, and the temperature is raised to $327^0C$, then the rate of emission of energy will become :

  1. $\dfrac{3}{8}E$
  2. $\dfrac{81}{16}E$
  3. $\dfrac{9}{16}E$
  4. $\dfrac{81}{64}E$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Let $A _1=32$ as given.
Let $A _2$ be the area when length and breadth are reduced by half. Thus the area will be $\dfrac {1}{4}$th of $A _1$
$ \therefore A _2=\dfrac {1}{4} *32=8$
Given $T _1={127}^0C={400}^0K$
Given $T _2={327}^0C={600}^0K$
From Stefan's law $E=\sigma AT^4$
$ \therefore\dfrac{E _1}{E _2}= \dfrac {A _1{T _1}^4}{A _2{T _2}^4}=\dfrac {32*(400)^4}{8*(600)^4}=\dfrac{64}{81}$
$ \therefore \dfrac{E _2}{E _1}= \dfrac{81}{64}$
Multiple choice thermodynamic principles of metallurgy general principles and processes of isolation of elements metallurgy chemistry

Hardened steel on heating in the range of $220^{\circ} C$ to $330^{\circ}C$ and on slow cooling gives :

  1. hard steel

  2. Brittle steel

  3. hard and brittle steel

  4. hard and tough steel
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Hardened steel on heating in the range of $220^{\circ} C$ to $330^{\circ}C$ and on slow cooling gives Hard and tough steel.

Multiple choice

What is the relationship between the critical temperature and the energy gap of a superconductor?

  1. They are directly proportional.

  2. They are inversely proportional.

  3. They are unrelated.

  4. The relationship depends on the specific superconductor.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The critical temperature and the energy gap of a superconductor are directly proportional.

Multiple choice

A metal rod of length (L) is heated at one end so that the temperature at a distance (x) from the heated end is given by the function (T(x) = 100 - 20x). What is the rate of heat flow through the rod at a distance (x = 2) meters from the heated end?

  1. \(-40\) W
  2. \(-20\) W
  3. \(20\) W
  4. \(40\) W
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The rate of heat flow through the rod is given by the equation (Q = -kA\frac{dT}{dx}), where (k) is the thermal conductivity of the rod, (A) is the cross-sectional area of the rod, and (\frac{dT}{dx}) is the temperature gradient. In this case, (k = 200) W/(m K), (A = 0.01) m^2, and (\frac{dT}{dx} = -20) K/m. So, the rate of heat flow at (x = 2) meters is (Q = -200(0.01)(-20) = -40) W.

Multiple choice

What is the rate of heat transfer by conduction?

  1. Q = kA(ΔT/L)

  2. Q = kA(ΔT/t)

  3. Q = kA(ΔT/x)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The rate of heat transfer by conduction is given by the equation Q = kA(ΔT/L), where Q is the amount of heat transferred, k is the thermal conductivity of the material, A is the area of contact between the two objects, ΔT is the temperature difference between the two objects, and L is the distance between the two objects.

Multiple choice

What is the rate of heat transfer by radiation?

  1. Q = σA(T^4)

  2. Q = σA(T^3)

  3. Q = σA(T^2)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The rate of heat transfer by radiation is given by the equation Q = σA(T^4), where Q is the amount of heat transferred, σ is the Stefan-Boltzmann constant, A is the surface area of the object, and T is the absolute temperature of the object.

Multiple choice

What is the formula for calculating the heat capacity of an object?

  1. C = m * c

  2. C = Q / ΔT

  3. C = Q * ΔT
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The formula for calculating the heat capacity of an object is C = m * c, where C is the heat capacity, m is the mass of the object, and c is the specific heat capacity of the material.

Multiple choice

What is the formula for calculating the specific heat capacity of a substance?

  1. c = Q / (m * ΔT)

  2. c = Q * ΔT

  3. c = Q / ΔT
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The formula for calculating the specific heat capacity of a substance is c = Q / (m * ΔT), where c is the specific heat capacity, Q is the amount of heat transferred, m is the mass of the substance, and ΔT is the change in temperature.

Multiple choice

Which thermodynamic relation expresses the relationship between the Joule-Thomson coefficient and the heat capacity at constant pressure?

  1. Joule-Thomson coefficient = heat capacity at constant pressure / temperature

  2. Joule-Thomson coefficient = heat capacity at constant pressure * temperature

  3. Joule-Thomson coefficient = heat capacity at constant pressure - temperature

  4. Joule-Thomson coefficient = heat capacity at constant pressure + temperature
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The Joule-Thomson coefficient is defined as the rate of change of temperature with pressure at constant enthalpy and is related to the heat capacity at constant pressure.

Multiple choice

What is the relationship between the inversion temperature and the van der Waals constants?

  1. Inversion temperature = 2a / Rb

  2. Inversion temperature = Rb / 2a

  3. Inversion temperature = 2a * Rb

  4. Inversion temperature = Rb + 2a
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The inversion temperature for a van der Waals gas is given by the equation: inversion temperature = 2a / Rb.